There are no bounded linear functional on $L^p(\mathbb{R})$ if $0Consider $L^p(\mathbb{R})$ where $0<p<1$. 
Why there are no bounded linear functional on  $L^p(\mathbb{R})$?
i.e.
If $l$ is linear functional $l:L^p(\mathbb{R})\to \mathbb{C}$ such that $|l(f)|\leq M|f|_{p}$ for all $f$ and some $M>0$, then $l=0$.
Why?
Hint: Let $F(x)=l(\mathcal{X}_{[0,x]})$,  and consider $F(x)-F(y)$
 A: This question surely was already answered here. In the case $0 <p <1$ the usual definition of the $L^p$-norm is not a norm, because the $\Delta$-inequality isn't satisfied. If $0< p <1$ the space $L^p(\mathbb{R})$ is only a metric space with invariant metric given by
$$\|f\|_p := \int |f|^p  d \lambda.$$
If $l$ is a continuous functional, we know that $|l(f)| < 1$ for $f \in B_\delta(0)$ and some $\delta >0$. By rescaling and using linearity, we see that $$|l(f)| \le M \|f\|_p^{1/p}$$
with $M= \delta^{-1}$. Now decompose $[a,b]$ in $n$ disjoint intervals $I_1,\ldots,I_n$ with lenght $(b-a)/n$. Then we have
$$|l(1_{[a,b]})| \le \sum_{k=1}^n |l(1_{I_i})| \le nM \|l(1_{I_i})\|_p = (b-a)^{1/p} n^{1-1/p}.$$
Since $0<p<1$, we see that $1-1/p$ is negative. Thus, letting $n \rightarrow \infty$, we get $l(1_{[a,b]}) =0$. Since all linear combinations of such intervals are dense in $L^p$ (also for $0<p<1$), we see that $l$ is trivial.
A: Suppose $\phi\in (L_p(\mathbb{R},\lambda))^*$ where $\lambda$ is the Lebesgue measure on $\mathscr{B}(\mathbb{R})$. If $\phi\neq0$, then there is $f\in L_p$ such that $\phi(f)\neq0$. Then either $\phi(f_+)>0$ or $\phi(f_-)>0$. This means that without loss of generality, we may assume that $f\geq0$ and $\phi(f)\leq1$. Define
$$F(x)=\int^x_{-\infty}f^p(t)\,dt$$
$F$ is a continuous function such that $\lim_{x\rightarrow-\infty}F(x)=0$ and $F(+\infty):=\lim_{x\rightarrow\infty}F(x)=\int f^p(t)\,dt$. Hence there is a point $x_1$ such that $F(x_1)=\frac12 F(+\infty)$. Define
\begin{align}
g_1 &=f\mathbb{1}_{(-\infty,x_1]},\qquad g_2 &=f\mathbb{1}_{(x_1,\infty)}
\end{align}
Clearly $f^p=g^p_1+g^p_2$. Since $\phi(g_1)+\phi(g_2)=\phi(f)\geq1$, then either $\phi(g_1)\geq \frac12$ or $\phi(g_2)\geq\frac12$. Choose $g\in \{g_1,g_2\}$ such that $\phi(g)\geq\frac12$, and define $f_1=2g$. Observe that $\phi(f_1)\geq1$ and
$$d(f_1,0)=2^p\int g=2^{p-1}\int f$$
Applying the same argument to $f_1$ in place of $f$, one obtains a function $f_2$ such that  $\phi(f_2)\geq1$ and
$$d(f_2,0)=2^{p-1}\int f^p_1=2^{n(p-1)}\int f^p$$
Continue this way add infinitum, we obtain a sequence $(f_n:n\in\mathbb{Z}_+)$ such that

*

*$d(f_n,0)=2^{(p-1)n}\int f$

*$\phi(f_n)\geq1$.

Since $0<p<1$, $d(f_n,0)\xrightarrow{n\rightarrow\infty}0$. The continuity of $\phi$ then would imply that $\phi(f_n)\xrightarrow{n\rightarrow\infty}0$; this however contradicts (2). Consequently $\phi=0$.
